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Find politeness of a number in C++
In this problem, we are given a positive integer N. Our task is to find the politeness of a number.
Polite Number is a number which can be expressed as a sum of two or more consecutive numbers.
Politeness of a number is defined as the number of ways the number can be expressed as sum of consecutive integers.
Let’s take an example to understand the problem,
Input
n = 5
Output
1
Explanation
2 + 3 = 5, is the only consecutive sum.
Solution Approach
A simple solution to the problem is to check all consecutive numbers till N and if their sum is equal to N, increase count which is the politeness of the number.
This solution is not efficient but a complicated but efficient solution is using factorisation. Using the formula for politeness that happens to be product of count of odd factors i.e.
If the number is represented as N = ax * by * cz… Politeness = [(x + 1)*(y +1)*(z + 1)... ] - 1
Program to illustrate the working of our solution,
Example
#include <iostream>
using namespace std;
int calcPolitenessNumber(int n){
int politeness = 1;
while (n % 2 == 0)
n /= 2;
for (int i = 3; i * i <= n; i += 2) {
int divCount = 0;
while (n % i == 0) {
n /= i;
++divCount;
}
politeness *= divCount + 1;
}
if (n > 2)
politeness *= 2;
return (politeness - 1);
}
int main(){
int n = 13;
cout<<"Politeness of "<<n<<" is "<<calcPolitenessNumber(n);
return 0;
}Output
Politeness of 13 is 1
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